Question

If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda }$ and the plane $x+2y+3z=4$ is ${\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right),$ then $\lambda$ equals

• A. $\frac{3}{2}$
• B. $\frac{2}{5}$
• C. $\frac{5}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

If $\theta$ be the angle between the given line and plane, then
$\sin \theta =\frac{1\times 1+2\times 2+\lambda \times 3}{\sqrt{1^2+2^2+{\lambda }^2}.\sqrt{1^2+2^2+3^2}}$
$=\frac{5+3\lambda }{\sqrt{14}.\sqrt{5+{\lambda }^2}}$
$\Rightarrow \cos \theta =\sqrt{1-\frac{{\left(5+2\lambda \right)}^2}{14\left(5+{\lambda }^2\right)}}$
$\therefore \theta ={\cos }^{-1}\sqrt{1-\frac{{\left(5+3\lambda \right)}^2}{14\left(5+{\lambda }^2\right)}}$
But it is given that $\theta ={\cos }^{-1}\sqrt{\frac{5}{14}}$
$\therefore \sqrt{1-\frac{{\left(5+3\lambda \right)}^2}{14\left(5+{\lambda }^2\right)}}=\sqrt{\frac{5}{14}}$
$\Rightarrow \lambda =\frac{2}{3}$